To find the center and radius of the circle, we'll need to rewrite the given equation in the standard form of a circle's equation:

$(x - h)^2 + (y - k)^2 = r^2$

where $(h, k)$ is the center of the circle, and $r$ is the radius. The given equation is:

$x^2 - 4x + y^2 + 5y - 17 = 0$

Step 1: Rearrange and Complete the Square

1. Group the $x$ and $y$ terms: $(x^2 - 4x) + (y^2 + 5y) = 17$

2. Complete the square for the $x$-terms:

   • Take half of the coefficient of $x$, which is $-4$, divide by 2 to get $-2$, and square it: $(-2)^2 = 4$.
   • Add and subtract 4 inside the equation.

   $(x^2 - 4x + 4) - 4$

   becomes

   $(x - 2)^2 - 4$

3. Complete the square for the $y$-terms:

   • Take half of the coefficient of $y$, which is $5$, divide by 2 to get $\frac{5}{2}$, and square it: $\left(\frac{5}{2}\right)^2 = \frac{25}{4}$.
   • Add and subtract $\frac{25}{4}$ inside the equation.

   $(y^2 + 5y + \frac{25}{4}) - \frac{25}{4}$

   becomes

   $(y + \frac{5}{2})^2 - \frac{25}{4}$

Step 2: Combine Everything Together

The equation now looks like:

$(x - 2)^2 - 4 + (y + \frac{5}{2})^2 - \frac{25}{4} = 17$

Combine the constants on the right side:

$(x - 2)^2 + (y + \frac{5}{2})^2 = 17 + 4 + \frac{25}{4}$

$(x - 2)^2 + (y + \frac{5}{2})^2 = \frac{68}{4} + \frac{25}{4}$

$(x - 2)^2 + (y + \frac{5}{2})^2 = \frac{93}{4}$

Step 3: Identify the Center and Radius

The standard form of the equation now is:

$(x - 2)^2 + (y + \frac{5}{2})^2 = \left(\frac{\sqrt{93}}{2}\right)^2$

Thus:

• Center: $(h, k) = (2, -\frac{5}{2})$
• Radius: $r = \frac{\sqrt{93}}{2}$

Final Answer

• Center: $(2, -\frac{5}{2})$
• Radius: $\frac{\sqrt{93}}{2}$

Would you like more details or have any questions about this process?

Here are 5 related questions:

1. How do you complete the square for quadratic expressions?
2. What is the general form of a circle’s equation?
3. How can you derive the radius from the standard form of a circle?
4. Can a circle have a negative radius? Why or why not?
5. What happens if a circle equation has coefficients other than 1 for $x^2$ and $y^2$?

Tip: Completing the square is a useful technique for converting quadratic expressions into a form that is easier to analyze or graph.